The monthly demand function for a product sold by a monopoly is
p = 3750 − 1/3x^2 dollars, and the average cost is C = 1000 + 70x + 3x^2
dollars. Production is limited to 1000 units and x is in hundreds of units.
(a) Find the quantity that will give maximum profit.
(b) Find the maximum profit. (Round your answer to the nearest cent.)
To find the quantity that will give maximum profit, we need to maximize the profit function, which is given by the formula:
Profit = Revenue - Cost
Revenue is calculated by multiplying the price (p) by the quantity (x), and cost is given by the average cost function (C).
(a) Find the quantity that will give maximum profit:
To find the quantity, we need to differentiate the profit function with respect to x and set it equal to zero.
Profit = Revenue - Cost
Profit = (p * x) - C
Differentiate the profit function with respect to x:
d(Profit)/dx = (dp/dx) * x + p - d(C)/dx
We already have the demand function p = 3750 - (1/3)x^2, the average cost function C = 1000 + 70x + 3x^2, but we still need to find dp/dx and d(C)/dx.
Differentiate the demand function with respect to x to find dp/dx:
dp/dx = -(2/3)x
Differentiate the average cost function C with respect to x to find d(C)/dx:
d(C)/dx = 70 + 6x
Now substitute the values we have into the derivative of the profit function:
d(Profit)/dx = (-(2/3)x) * x + (3750 - (1/3)x^2) - (70 + 6x)
Simplify the equation:
-(2/3)x^2 + (3750 - (1/3)x^2) - (70 + 6x) = 0
Combine like terms:
3750 - (1/3)x^2 - 70 - 6x - (2/3)x^2 = 0
Simplify further:
(2/3)x^2 + (2/3)x - 3680 = 0
To solve the quadratic equation, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
In this case, a = 2/3, b = 2/3, and c = -3680. Calculating the values:
x = [-(2/3) ± sqrt((2/3)^2 - 4 * (2/3) * (-3680))] / (2 * (2/3))
Simplifying:
x = [-(2/3) ± sqrt(4/9 + 29440/3)] / (4/3)
x = [-(2/3) ± sqrt(4/9 + 88213.333)] / (4/3)
x = [-(2/3) ± sqrt(88218.667)] / (4/3)
Now we have two possible values for x. We will choose the one that is within the production limit of 1000 units (x in hundreds of units) and positive:
x = [-(2/3) + sqrt(88218.667)] / (4/3)
Calculating this formula will give us the quantity that will give maximum profit.
(b) Find the maximum profit:
Once we have the quantity, we can use the demand function and the cost function to calculate the price and cost at that quantity. Then, we can calculate the profit using the formula:
Profit = (p * x) - C
where p is the price and x is the quantity. Calculate the profit and round the answer to the nearest cent to find the maximum profit.
To find the quantity that will give the maximum profit, we need to determine the level of production (quantity) at which the profit is maximized. The profit function is given by subtracting the cost function from the revenue function.
The revenue function is calculated by multiplying the price (p) by the quantity (x):
Revenue (R) = p * x
The cost function is given as:
C = 1000 + 70x + 3x^2
Therefore, the profit function (P) can be calculated as:
P = R - C
Substituting the value of p from the demand function into the revenue function, we have:
R = (3750 - (1/3)x^2) * x
R = 3750x - (1/3)x^3
Substituting the value of C into the profit function, we have:
P = (3750x - (1/3)x^3) - (1000 + 70x + 3x^2)
P = 3750x - (1/3)x^3 - 1000 - 70x - 3x^2
To find the quantity that will give maximum profit, we need to find the value of x that maximizes the profit function P. We can do this by taking the derivative of P with respect to x, setting it equal to zero, and solving for x.
dP/dx = 3750 - x^2 - 6x
0 = 3750 - x^2 - 6x
x^2 + 6x - 3750 = 0
To solve this equation, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
x = (-6 ± sqrt(6^2 - 4 * 1 * -3750)) / (2 * 1)
x = (-6 ± sqrt(36 + 15000)) / 2
x = (-6 ± sqrt(15036)) / 2
x = (-6 ± 122.68) / 2
Simplifying further, we have:
x = (-6 + 122.68) / 2 = 116.68 / 2 = 58.34
x = (-6 - 122.68) / 2 = -128.68 / 2 = -64.34
Since production cannot be negative, we discard the negative value of x. Therefore, the quantity that will give the maximum profit is x = 58.34.
To find the maximum profit, we substitute this value of x back into the profit function:
P = 3750x - (1/3)x^3 - 1000 - 70x - 3x^2
P = 3750(58.34) - (1/3)(58.34)^3 - 1000 - 70(58.34) - 3(58.34)^2
Calculating this expression will give us the maximum profit.